A topology can be enormous. The standard topology on the line \mathbb{R} is the collection of all open sets — every open interval, every union of intervals, every wild open set with countably many gaps. There is no hope of listing them, and no hope of checking a claim "for every open set" by brute force. Yet a topology on \mathbb{R} can be pinned down completely by a tiny, tidy family: the open intervals (a, b). Every open set — however baroque — is just a union of these. Name the intervals and you have named the whole topology.

This is the same economy that runs through all of mathematics. You do not describe a vector space by listing its infinitely many vectors; you give a basis and take linear combinations. You do not describe a group by writing out its multiplication table; you give generators. A topology gets the same treatment. A small generating family that unfolds into the whole topology by taking unions is called a basis, and an even leaner family that unfolds by taking finite intersections and then unions is called a subbasis. This page is about those two levers — how they generate a topology, which collections are allowed to be a basis, and how they let you compare topologies at a glance.

Throughout, X is a set and a topology \tau on it is a collection of subsets (the open sets) closed under arbitrary unions and finite intersections, with \varnothing and X themselves open. Everything below is machinery for describing such a \tau without writing all of it down.

What a basis is

Fix a topology \tau on X. A subcollection \mathcal{B} \subseteq \tau of open sets is a basis for \tau if every open set is a union of members of \mathcal{B}:

There is an equivalent, more usable phrasing — the one you actually check in proofs. A collection \mathcal{B} of open sets is a basis for \tau exactly when it satisfies the local condition:

The two versions say the same thing. If every point of U has a basis set squeezed between it and U, then U is the union of all those little basis sets; conversely a union of basis sets obviously houses a basis set around each of its points. Think of the basis elements as the "standard neighbourhoods": to be open is precisely to contain a standard neighbourhood around each of your points.

The prototype: on \mathbb{R}, the open intervals \mathcal{B} = \{\, (a, b) : a < b \,\} form a basis for the standard topology. A set U is open iff around each of its points sits a little interval inside U — the familiar picture. More generally, in any metric space (X, d), the open balls B(x, r) = \{\, y : d(x, y) < r \,\} form a basis for the metric topology — a set is open iff it contains a ball around each of its points, which is the very definition of an open set in a metric space.

Which collections may be a basis? The two conditions

Here is the striking part. You can start with a raw collection \mathcal{B} of subsets of X — no topology given in advance — and ask: is there a topology for which this is a basis? There is exactly one candidate (unions of members of \mathcal{B}), and it works iff \mathcal{B} passes two tests.

Read the conditions as exactly what the topology axioms demand of a generating family. Condition (B1) is there so that X itself comes out open (it must be a union of basis sets, so the basis had better reach every point). Condition (B2) is there so that the union topology is closed under intersection: if B_1 and B_2 are open, their intersection must be a union of basis sets too, and the only way to guarantee that is to fit a basis set into B_1 \cap B_2 around every point of the overlap. Note (B2) does not ask that B_1 \cap B_2 itself be a basis element — only that it be coverable from inside by basis elements.

The open intervals pass both effortlessly: they cover \mathbb{R}, and the intersection of two intervals is again an interval (or empty), so (B2) holds with B_3 = B_1 \cap B_2 itself. Open balls pass too — that is Worked example 2 below, and it is the one place (B2) needs a genuinely non-trivial choice of B_3.

The picture: covering and the intersection condition

In the plane, take the open discs as your basis. An open set U is painted by laying overlapping discs across it until every point is inside one — that is condition (B1) at work, U as a union of discs. The subtle part is what happens where two discs overlap: pick a point x in B_1 \cap B_2, and condition (B2) demands a third disc B_3, entirely inside the lens-shaped overlap, still containing x. Step through the figure to watch the cover build and the nested disc appear.

The overlap B_1 \cap B_2 — the lens — is not a disc, so it is not itself a basis element. But (B2) does not ask it to be. It only asks that we can slip a genuine disc B_3 around x inside the lens, and we always can (shrink the radius until the disc clears both boundaries). That single move is what makes the discs a legal basis, and it is exactly the balls-are-a-basis argument dressed in pictures.

Worked example 1: generating a topology from a small basis

Let X = \{a, b, c\} and take the collection

Check (B1). Does \mathcal{B} cover X? Yes — a \in \{a\} and b, c \in \{b, c\}. Every point is housed.

Check (B2). Run through the pairwise intersections. \{a\} \cap \{b,c\} = \varnothing (nothing to check — no point x to serve). \{a\} \cap \{a,b,c\} = \{a\}, which is a basis element, so take B_3 = \{a\}. Similarly \{b,c\} \cap \{a,b,c\} = \{b,c\} \in \mathcal{B}. Every overlap contains a basis set around each of its points, so (B2) holds.

Generate. The topology is all unions of members (including the empty union):

Take unions of the three basis sets and you get nothing new — \{a\} \cup \{b,c\} = X is already listed. So this four-set collection is the generated topology, and you can verify by hand it is closed under unions and intersections. Notice how a three-element basis produced the topology outright; on a small space the basis is barely smaller than \tau, but on \mathbb{R} the compression from "all intervals" to "all open sets" is total.

Worked example 2: open balls really are a basis

On a metric space (X, d), let \mathcal{B} = \{\, B(x, r) : x \in X,\ r > 0 \,\} be the collection of all open balls. We show it satisfies (B1) and (B2), so it generates a topology — the metric topology.

(B1). Every point x lies in a ball around itself, e.g. x \in B(x, 1). The balls cover X.

(B2). Suppose x \in B(p, r) \cap B(q, s). Then d(p, x) < r and d(q, x) < s, so the two "slacks" r - d(p, x) and s - d(q, x) are both positive. Put

We claim B(x, \varepsilon) \subseteq B(p, r) \cap B(q, s). Take any y \in B(x, \varepsilon). By the triangle inequality,

so y \in B(p, r); the identical estimate with q, s gives y \in B(q, s). Hence x \in B(x, \varepsilon) \subseteq B(p, r) \cap B(q, s), which is precisely (B2) with B_3 = B(x, \varepsilon). The balls are a basis. This is the lens-and-nested-disc picture made rigorous: \varepsilon is exactly how much you shrink the radius to clear both boundaries.

Subbases: generating from even less

A basis already asks for something — the intersection condition (B2). A subbasis asks for essentially nothing, and repairs the deficit by doing two operations instead of one.

Why does the finite-intersection family automatically satisfy (B2)? Because it is closed under finite intersection by construction: if B_1 and B_2 are each a finite intersection of subbasis sets, then B_1 \cap B_2 is one too, so it is itself a basis element and you may take B_3 = B_1 \cap B_2. That is the whole trick: a subbasis need not close up under intersection because forming the basis does the closing for you. The only requirement left on \mathcal{S} is that it cover X, and even that is often waved through by counting the empty intersection as X.

"Smallest topology containing \mathcal{S}" deserves a second look. Any topology \tau that contains \mathcal{S} must, being a topology, also contain all finite intersections of members of \mathcal{S} and then all unions of those — that is, it must contain the generated topology. So the generated topology is a subset of every topology containing \mathcal{S}: it is the coarsest, the smallest, the intersection of them all. A subbasis is thus the most economical possible way to specify a topology — hand over any cover of X and the machine builds the leanest topology that makes all those sets open.

Worked example 3: rays as a subbasis for the line

Take X = \mathbb{R} and the collection of all open rays:

These cover \mathbb{R} (every real is in some (-\infty, b)), so \mathcal{S} is a legitimate subbasis. Now form finite intersections. A ray with a ray of the same type gives another ray ((-\infty, 3) \cap (-\infty, 7) = (-\infty, 3)). The interesting intersection is opposite types:

So the finite intersections of rays are exactly the open intervals (a, b) (plus the rays themselves, and \varnothing when a \ge b). That is precisely the standard basis of intervals from the opening card — whose unions give the standard topology. Conclusion: the two families of rays form a subbasis for the standard topology on \mathbb{R}. Two crude ingredients — "everything below b" and "everything above a" — intersect to produce every interval, and the whole real topology unfolds from there.

Comparing topologies through their bases

Two topologies on the same set can be compared: \tau' is finer than \tau (and \tau coarser than \tau') when \tau \subseteq \tau' — every set open in the coarse one is open in the fine one. "Finer" means more open sets, more distinctions, smaller neighbourhoods. Bases turn this set-containment question into a purely local check.

The classic illustration lives on \mathbb{R}. The standard topology has basis the open intervals (a, b). The lower-limit topology \mathbb{R}_\ell has basis the half-open intervals [a, b). Is the lower-limit topology finer? Given a standard basis set (a, b) and a point x in it, the half-open set [x, b) satisfies x \in [x, b) \subseteq (a, b) — yes, it fits. So \mathbb{R}_\ell is finer. And it is strictly finer: the set [0, 1) is open in \mathbb{R}_\ell but not in the standard topology, because no open interval around 0 fits inside [0, 1) (any interval around 0 spills to the left of 0). The basis criterion caught the refinement with one line of interval-fitting, no need to reason about all open sets at once.

The three levers, in one view

Stand back and the whole page is a hierarchy of generating families, each one leaner and each one requiring one more operation to unfold into the full topology.

• Topology \tau: the complete list of open sets. Closed under arbitrary unions and finite intersections. Usually far too big to write down.
• Basis \mathcal{B}: take arbitrary unions to recover \tau. Must satisfy (B1) cover and (B2) intersection.
• Subbasis \mathcal{S}: take finite intersections, then arbitrary unions to recover \tau. Must only cover X. Generates the smallest topology containing it.

On \mathbb{R} the chain is vivid: the two families of rays (subbasis) intersect to the open intervals (basis) which union to every open set (topology). Each arrow discards structure you never needed to store. That is the payoff — you specify a topology by its cheapest generators and let unions and intersections do the rest.